A076312 a(n) = floor(n/10) + 2*(n mod 10).

0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 7, 9, 11, 13, 15, 17, 19, 21

Offset: 0

Author

Reinhard Zumkeller, Oct 06 2002

Keywords

• nonn

Comments

Delete the last digit from n and add twice this digit to the shortened number. - N. J. A. Sloane, May 25 2019

(n==0 modulo 19) iff (a(n)==0 modulo 19); applied recursively, this property provides a useful test for divisibility by 19.

Examples

			26468 is not a multiple of 19, as 26468 -> 2646+2*8=2662 -> 266+2*2=270 -> 27+2*0=27=19*1+8, therefore the answer is NO.
Is 12882 divisible by 19? 12882 -> 1288+2*2=1292 -> 129+2*2=133 -> 13+2*3=19, therefore the answer is YES.
		

References

• Erdős, Paul, and János Surányi. Topics in the Theory of Numbers. New York: Springer, 2003. Problem 6, page 3.
• Karl Menninger, Rechenkniffe, Vandenhoeck & Ruprecht in Goettingen (1961), 79A.

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
• Eric Weisstein's World of Mathematics, Divisibility Tests.
• Wikipedia, Divisibility rule
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,1,-1).

Crossrefs

Cf. A008601, A076309, A076310, A076311.

Programs

Haskell

a076312 n =  n' + 2 * m where (n', m) = divMod n 10
-- Reinhard Zumkeller, Jun 01 2013

Magma

[Floor(n/10) + 2*(n mod 10): n in [0..100]]; // Vincenzo Librandi, Mar 05 2020

Mathematica

f[n_]:=Module[{idn=IntegerDigits[n]},FromDigits[Most[idn]]+2idn[[-1]]]; Array[ f,80,0] (* Harvey P. Dale, Mar 01 2020 *)

Formula

G.f.: -x(17x^9-2-2x-2x^2-2x^3-2x^4-2x^5-2x^6-2x^7-2x^8)/((1-x)^2(1+x)(x^4+x^3+x^2+x+1)(x^4-x^3+x^2-x+1)). a(n)=A059995(n)+2*A010879(n). [R. J. Mathar, Jan 24 2009]